If $3 \mathrm{f}(x)-\mathrm{f}\left(\frac{1}{x}\right)=8 \log _2 x^3, x>0$, then $\mathrm{f}(2),…

If $3 \mathrm{f}(x)-\mathrm{f}\left(\frac{1}{x}\right)=8 \log _2 x^3, x>0$, then $\mathrm{f}(2), \mathrm{f}(4)$ $\mathrm{f}(8)$ are in
  1. A.P.
  2. G.P.
  3. H.P.
  4. Arithmetico Geometric Progression.

Solution

$\begin{aligned} & 3 f(x)-f\left(\frac{1}{x}\right)=8 \log _2 x^3 \\ & \Rightarrow 3 \mathrm{f}\left(\frac{1}{x}\right)-\mathrm{f}(x)=8 \log _2\left(\frac{1}{x}\right)^3 . \end{aligned}$ From (i) and (ii), we get $\begin{aligned} & 8 \mathrm{f}(x)=24 \log _2 x^3+8 \log _2\left(\frac{1}{x}\right)^3 \\ & \Rightarrow 8 \mathrm{f}(x)=72 \log _2 x-24 \log _2 x \\ & \Rightarrow 8 \mathrm{f}(x)=48 \log _2 x \\ & \Rightarrow \mathrm{f}(x)=6 \log _2 x \end{aligned}$ $\begin{array}{ll}\therefore \quad & f^{\prime}(2)=6 \log _2 2=6 \\ & f^{\prime}(4)=6 \log _2 4=12 \\ & f^{(}(8)=6 \log _2 8=18 \\ \therefore \quad & f(2), f(4), f(8) \text { are in A.P. }\end{array}$

Asked in: MHT CET 2023 (13 May Shift 1)

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